IN THE ADJOINING FIGURE ,TRIANGLE ABC IS AN ISOSCELES TRIANGLE IN WHICH AB=AC.IF E AND F BE THE MIDPOINTS OF AC AND AB RESPECTIVELY,

PROVE THAT BE=CF.

Given: In ΔABC, AB = AC

and D and E are mid point of AB and AC

⇒ AD = BD = AE = CE  .....(1)

In ΔABC

∠ABC = ∠ACB  (Angles opposite to equal sides)

⇒ ∠DBC = ∠ECD  ......(2)

Now, In ΔDBC and ΔECB

DB = EC  (From (1))

∠DBC = ∠ECD  (From (2))

BC = CB  (Common)

So, ΔDBC ΔECB  (By SAS congruency criterion)

⇒ BE = CF  (CPCT)

  • -2

Hi.

AB = AC(given)

F and E are midpoints

This implies FB = EC

& angle FBC = angle ECB(angles opposite to equal sides are also equal)

In triangle CBE and triangle BCE

FB = EC(just proved)

CB = BC(common)

Angle FBC = Angle ECB(just proved)

Therefore by SAS congruence condition

Triangle BCF triangle BCE

(By the way sorry for not making a figure. I don't know how to make it on computer)

 

Hope my answer helps :) . Cheers!!!

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